x Level A Maths Ch 5 Trigonometry
PapaCambridge 180-250: 71 problems
° π θ α β γ
ʃ √ ≅ ≡ ≈ ≠ ⇒
cos−1
6⋅cos θ ⋅tanθ −3⋅cos θ + 4 tanθ −2 ⇔
√
3
()
a
b
Pshaded = |^
BD|+|DC|+|CB|
1
2
Continue with Orion problems---20241019
PapaCambridge 180-250: 71 problems
210
209
208—part (iii) is subtle! ---------- see third page of solution for easier example.
207
206
205
204
203
202
201 – calculator graphing – mistake in (c) (iii)?
200
199 - 20241108.1424 to 20241108.1438 ⇒ 14 minutes
198
197
196 – Ch 3 Coordinate geometry problem
195 – Ch 4 circular geometry problem
194
193 - AND PapaCambridge 222
192------- far better to do my hand, inking over existing graph, or by calculator
191
190
189
188-easy fundamental example
187—excellent graphing-shifting and use of TI84 color graphing
186
185
183. 9709_s21_qp_11 Q:11
182. 9709_s21_qp_11 Q:4
181.9709_m21_qp_12 Q:3
180. 9709_m22_qp_12 Q: 7
−b±√ b2−4 ac
2a
180. 9709_m22_qp_12 Q: 7
180. 9709_m22_qp_12 Q 7
Below is my solution in detail as well as a “proof” by graphing.
A TI-84 graph of
Y3 =
sin x+ 2cos x sin x−2 cos x
−
= LHS
sin x−2 cos x sin x +2 cos x
and
Y9 =
4
4
=
= RHS
2
2
5 cos θ−4
5 cos x−4
on the interval (o , π ) = (o ,180 ° ) seem to show that these two functions are equal:
Compare these two screen shots from the TI-84 showing
sin x+ 2cos x sin x−2 cos x
−
in black:
sin x−2 cos x sin x +2 cos x
4
in red and, underneath is, hiding
2
5 cos x−4
Here are all the steps we used in our proof:
LHS =
(sin θ+2 cos θ)⋅(cos θ+ 2sin θ)−(sin θ−2cos θ)⋅(cos θ−2 sin θ)
sin θ+2 cos θ sin θ−2 cos θ
−
=
cos θ−2sin θ cos θ+2 sin θ
(cos θ−2 sin θ)⋅(cos θ+2 sin θ)
2
2
2
2
=
sin θ cos θ+2 sin θ+2 cos θ+ 4 cos θ sin θ−[sin θ cos θ−2sin θ−2 cos θ+4 sin θ cos θ]
(cos θ−2sin θ)⋅(cos θ+2 sin θ)
=
2
2
4
4
4 sin θ+ 4 cos θ+ 0 sinθ cos θ
=
=
= RHS
2
2
2
2
2
cos θ−4 sin θ
5 cos θ−4
cos θ−4⋅(1−cos θ)
with annotations:
181.9709_m21_qp_12 Q:3
182. 9709_s21_qp_11 Q:4
183. 9709_s21_qp_11 Q:11
Or, using the equation editor
2
LHS =
2
2
2
2
1−2 sin θ
1−sin θ−sin θ
1−sin θ
sin θ
=
=
= 1−tan2 θ = RHS
2
2
2
2
1−sin θ
1−sin θ
cos θ
cos θ
Evaluation on the TI-89 yields 35.264 389 682 8° ≈ 35.3°≈ 35° for one solution.
There is a second solution at 180° – 35.264 389 682 8° ≈ 144.7°.
A TI-84 graph of
Y 5 = 1−tan2 x = LHS
and
Y 6 = 2 tan 4 x = RHS
on the interval (0 , π) = (0 , 180 °) seem to show that these two functions’ graphs cross at about
θ = 0.6 and θ = π - 0.6 = 2.5, or θ = 35° θ = 35° and θ = 145°. The more precise values indicated above are
35.264 389 682 8° ≈ 35.3° and 180° – 35.264 389 682 8° = 144.736 610 317 2° ≈ 144.7°.
Note that, if you cannot prove an equality is true, you can check it for some easy values of θ.
Nothing is easier than θ = 0, which yields
LHS =
2
1
1−2 sin 0
=
= 1
2
1
1−sin 0
and RHS = 1−tan2 0 = 1
This doesn’t PROVE that LHS = RHS , but certainly SUPPORTS equality.
A more daring test value for θ would be any multiple of
multiple of
π
. And still more daring, a
2
π
π
or a multiple of
.
4
6
For example, if θ =
π
, then
4
π
1
2
1−2⋅( )
1−2 sin ( √ )
(
)
4
2
2
0
LHS =
=
=
=
= 0
1
π
1
2
√
1−sin ( )
1−( )
1−sin ( )
2
4
2
2
1−2 sin2
2
2
2
and RHS = 1−tan
2
π
= 1−12 = 0
4
This still doesn’t PROVE that LHS = RHS , but certainly gives us more confidence it is true.
185. 9709_s21_qp_13 Q 4 20231106
reference symbols: Σ ∞ ∞ - ° → ◊ ◊ π α β γ θ ʃ √ ≅ ≡ ≈ ≠ ε
n
n
n(n+1)
a 11 a 12 a13
n
k
lim ∑ f (x k )⋅Δ x ⇒
∑
a 21 a 22 a 23
x
2
n→∞ k=1
1
(
) ()
μ σ χ
φ
Δx
'
f (x)
185
This is true if we show that
tan x +sin x
1+cos x
=
tan x−sin x
1−cos x
(
(
) = sin x+ cos x sin x = (sin x )⋅(1+cos x)
sin x−cos x sin x
(sin x)⋅(1−cos x)
)
sin x
sin x
cos x⋅
+ sin x
+sin x
cos x
cos x
tan x +sin x
LHS =
=
=
sin x
tan x−sin x
sin x
−sin x
cos x⋅
−sin x
cos x
cos x
=
Therefore
Since
1+cos x
= RHS
1−cos x
tan x +sin x
1+cos x
= k is equivalent to
= k
tan x−sin x
1−cos x
1+cos x
= k
1−cos x
1+cos x = k⋅(1−cos x) = k −k⋅cos x
and
cos x +k cos x = k −1 ⇒ cos x =
k−1
k +1
From (a), this is equivalent to
1+cos x
= 4
1−cos x
Using (b) this is equivalent to cos x =
4−1
3
=
4 +1
5
⇒
−1
x = cos
( 35 ) ≈ 0.927 295 218≈ 0.927
But on the interval (−π , π ) , another solution is x ≈ -0.927 295 218≈ -0.927 in the fourth quadrant.
A TI-84 graph of Y 3 =
tan x +sin x
= LHS and Y 9 = 4 = RHS on the interval (−π , π ) = (−180 ° ,180 °)
tan x−sin x
seem to show that these two functions are equal near ±0.927 295 218 :
186
NB: The proper way to attach this is to use ALGEBRA, NOT TRIGONOMETRY FIRST
0 = 6⋅cos θ ⋅tanθ −3⋅cos θ + 4 tanθ −2 ⇒ 0 = 6⋅cos θ ⋅sin θ −3⋅cos 2θ +4 sinθ −2 cos θ does not help.
However, if we begin by factoring the quadrinomial, we get
0 = 6⋅cos θ ⋅tanθ −3⋅cos θ + 4 tanθ −2 ⇒ 0 = (2 tan θ −1)⋅(3⋅cos θ +2)
⇒
tanθ =
1
2
or cos θ =−
2
3
−1
⇒ θ =tan
( 12) or θ =cos (−23)
−1
On a TI89 these can be evaluated with 2 lines input or one-line input using a semi-colon:
Here is a TI84 graph of the original equation plotted from 0° to 264° traced to x = 132° near the 131.810° solution.
NB: 264 = 4 * 66.
187—excellent graphing-shifting and use of TI84 color graphing
Y 1=cos (x ) Y 3=cos(2 x) Y 4 =5 cos(2 x ) Y 5=5 cos (2 x)+3 on the interval [0 ,2 π ]
187 (cont)
Y 3=cos(2 x) Y 4 =5 cos( 2 x ) Y 5=5 cos (2 x)+3 on the interval [0 , π ]
on the interval [0 ,2 π ] :
6x
6x
has 3 solutions and Y 5=5 cos (2 x)+3 = 6−
has 2 solutions
Y 5=5 cos (2 x)+3 =
π
π
188-easy fundamental example
2⋅cos θ =7−
3
cos θ
2
⇒ 0 = 2⋅cos θ −7 cos θ +3 = (2⋅cos θ −1)⋅(cos θ −3)
−1
⇒ 2⋅cos θ −1 = 0 or cos θ −3 = 0 ⇒ θ =cos
( 12 ) or θ =cos 3 ⇒ θ =±60°
−1
189
(a)
tan x +cos x
= k
tan x−cos x
⇒
tan x +cos x = k⋅(tan x−cos x) ⇒ sin x +cos 2 x = k⋅(sin x−cos 2 x )
0 = (k + 1)⋅cos 2 x +(1−k ) sin x = (k + 1)⋅(1−sin 2 x)+(1−k ) sin x = (k + 1)⋅(1)−(k + 1)⋅sin 2 x +sin x−k⋅sin x
2
⇒ (k + 1)⋅sin x +(k−1)sin x−(k +1) = 0
(b) From (a) this is equivalent to
0 = 5⋅sin 2 x +3 sin x−5 ⇒ sin x =
⇒
x = sin−1
( −3±10√109 )
−b±√ b2−4 ac
−3± √ 9+100
−3± √ 109
=
=
2a
10
10
190
(a)
tan θ +3⋅sin θ +2
= 2 ⇒
tanθ −3 sin θ + 1
tanθ +3⋅sin θ + 2 = 2⋅(tan θ −3 sin θ +1)
⇒ sin θ +3⋅sin⋅cos θ +2⋅cos θ = 2⋅sin θ −6⋅sin θ ⋅cos θ + 2⋅cos θ
⇒ 0 = sin θ −9⋅sin⋅cos θ = sin θ ⋅(1−9⋅cos θ ) ⇒ sin θ =0 or cos θ =
−1
⇒ θ =sin−1 0 = 0 or θ =cos
( 19 ) ≈ 83.6 °
1
9
191
(a) 0 = 3 tan 2 x−5 tan x−2 = (3 tan x +1)⋅(tan x−2) ⇒
⇒
−1
−1
x=tan 2 ≈ 63.4 ° or x=tan
(b) 0 = 3 tan 2 x−5 tan x +k = ⇒
1
≈ −18.4 ° ⇒
3
tan x =
no solutions iff 25−12 k <0 ⇒ k >
tan x=2 or tan x=−
1
3
x=63.4 ° is only solution on [0 ° , 180 °]
−b±√ b2−4 ac
5±√ 25−12k
=
2a
6
25
12
THIS IS A DIFFICULT POINT!
25
5± √ 25−12 k
Since for k <
the solution pairs of tan−1
repeat with period 180°, we must have k be such
12
6
that both 0° and 180°are solutions. This occurs of k =0 making
(
tan−1
)
10
k
5± √ 25
= tan (
yielding tan 0 =0° and 180°and tan
≈ 59.0 °
( 5± √25−12
)
)
6
6
6
−1
−1
−1
The roots come in pairs exactly 180° apart due to the period of 180° the tangent function.
192------- far better to do my hand, inking over existing graph, or by calculator
3
1
Y 1= ⋅cos 2 x+
2
2
3
3
Y 2= ⋅cos 2 x +
2
2
3
1
Y 1= ⋅cos 2 x+
2
2
3
1
Y 3=−Y 1=− ⋅cos 2 x −
2
2
3
3
Y 2= ⋅cos 2 x +
2
2
3
1
Y 3=−Y 1=− ⋅cos 2 x −
2
2
19 - AND PapaCambridge 222
(a) LHS =
=
2
2
(1+sin θ )⋅(1+sin θ )+(cos θ )⋅(cos θ )
1+ sinθ
cos θ
1+ 2sin θ +sin θ +cos θ
+
=
=
cos θ
1+ sinθ
(cos θ )⋅(1+sin θ )
(cos θ )⋅(1+sin θ )
2⋅(1+ sinθ )
2+2 sin θ
2
=
=
= RHS
cos
θ
(cos
θ
)⋅(1+
sin
θ
)
(cos θ )⋅(1+ sinθ )
(b)
1+ sinθ
cos θ
3
+
=
cos θ
1+ sinθ
sinθ
−1
⇒ θ = tan
⇒
2
3
=
cos θ
sinθ
2 sin θ = 3 cos θ
( 32 ) ⇒ θ ≈ 0.983 , 0.983+ π ≈ 0.983 , 4.124
194
(a) 3 cos θ =8 tan θ
⇒ 3 cos 2θ =8sin θ
⇒ 3⋅(1−sin 2θ )=8 sin θ
⇒ 0=3 sin2θ + 8sin θ −3
−8±10
−8±10
1
−b±√ b2−4 ac
−8± √ 64 +36
=
=
=
=
or 3
⇒ sin θ =
6
6
3
2a
6
⇒ θ = sin
−1
( 13 ) ≈ 19.47 °
195 – Ch 4 circular geometry problem
(a) By Law of Cosines |BC| =
2
π
2
√3
r⋅( π +3+ 3⋅√ 5−2⋅√ 3)
= |⏞
AB|+|AC|+|CB| = 2 r⋅ π +r + r⋅√ 5−2⋅√ 3 =
(6)
3
arc
(b) Pshaded
√
√(r +( 2r ) −2⋅r⋅2 r⋅cos 6 ) = r⋅ (5−4⋅ 2 ) = r⋅√5−2⋅√ 3
2
A sectorOAB −A Δ OCB =
r ⋅(2⋅π −3)
1
1
2
(2 r ) ⋅π − ⋅r⋅2 r⋅sin ( π ) =
2
6 2
6
6
196 – Ch 3 Coordinate geometry problem
(a) The range of Y 1=(2−3 sin(2 X))⋅( X≤π )⋅( X≥0) is [−1 ,5]
The range of Y 2=(−4 +6 sin(2 X))⋅( X≤π )⋅( X≥0) is [2,−10]
(b)
(c) Stretch of Y 1 by a factor of 2 and reflect about the x-axis to obtain Y 2 .
Then shift Y 2 π units to the left to obtain Y 3=(−4+6 sin (2⋅( X + π )))⋅(X ≤0)⋅( X≥−π )
197
(a) LHS =
=
tan θ ⋅(1−cos θ +1+cos θ )
tan θ
tanθ
2 tan θ
2 sinθ
+
=
=
=
2
2
1+ cos θ 1−cos θ
(1+cos θ )⋅(1−cos θ )
(1−cos θ )
cos θ ⋅(1−cos θ )
2 sin θ
2
=
= RHS
2
cos θ ⋅sin θ
cos θ ⋅sin θ
tan θ
tanθ
6
+
=
1+ cos θ 1−cos θ
tan θ
⇒
1
= 3⋅cos θ
cos θ
⇒
2
6
6⋅cos θ
=
=
cos θ ⋅sin θ
tan θ
sinθ
⇒ cos 2θ =
1
3
⇒
2
= 6⋅cos θ
cos θ
−1
√ 3 ≈ 54.7°, 125.3°
⇒ θ = cos ±
3
198
Y 1=cos (X ) , Y 2=2 cos ( X /2)+3
199 - 20241108.1424 to 20241108.1438 ⇒ 14 minutes
LHS =
2
2
sinθ ⋅(1+sin θ )−sin θ ⋅(1−sin θ )
sin θ
sin θ
2⋅sin θ
2⋅sin θ
−
=
=
=
= 2⋅tan2θ = RHS
2
2
1−sin θ 1+ sinθ
(1−sinθ )⋅(1+sin θ )
(1−sin θ )
cos θ
sin θ
sin θ
−
= 8 ⇒ 2⋅tan2θ = 8 ⇒
1−sin θ 1+ sinθ
⇒ θ = ±tan−1 2 ≈ 63.4°, 116.6°
2
tan θ = 4
⇒
tan θ = ±2
200
(a) LHS =
=
(
)(
) (
1
1
−tan x ⋅
+1 =
cos x
sin x
)(
1−sin x 1+sin x
⋅
cos x
sin x
)=(
2
1−sin x
cos x⋅sin x
)=(
2
cos x
cos x⋅sin x
)
cos x
1
=
= RHS
sin x
tan x
( cos1 x −tan x)⋅( sin1 x +1) = 2⋅tan x ⇒ tan1 x = 2⋅tan x ⇒ tan x = √ 12 ⇒ x = tan ( √ 12 ) ≈ 38.4°
2
2
3
−1
3
201 – calculator graphing – mistake in (c) (iii)?
(i) k =−3 0 solutions
(ii) k =1 2 solutions
(iii) k =3 2 solutions [mistake --- meant k = -1?]
(iv) k =−1 2 solutions
202
2
(a) 3⋅tan x+1=
⇒
2
tan x =
2
2
tan x
⇒ 3⋅tan4 x + tan2 x=2 ⇒ 0=3⋅tan4 x+ tan2 x−2
−1±5
2
−1±√ 1+24
=
= −1 or=
6
3
6
⇒
tan x = ±
√
2
3
⇒
x = ±tan−1
√
2
≈ 39.2°, 140.8°
3
203
(a) 0 = 3⋅sin2 2 θ +8 cos 2 θ = 3⋅(1−cos 2 2θ )+ 8 cos 2θ
⇒ 0 = 3 cos 2 2 θ −8 cos 2 θ −3 = (3 cos 2 θ +1)⋅(cos 2θ −3) ⇒ cos 2 θ =−
⇒ θ =
( ) ≈ 54.7°, 125.3°
1
1
−1
⋅cos −
2
3
1
3
(b) y = a+ tan bx ⇒ 0 = a+ tan
⇒
(−b⋅6 π ) and √ 3 = a ⇒ tan( −b⋅6 π ) = −√ 3
6
b⋅π
6
6 π
−1
= tan−1 √ 3 ⇒ b = π tan √ 3 = π ⋅π = ⋅ = 2
6
3
π 3
204
(i)
(
1
−tan x
cos x
) (
2
=
1−sin x
cos x
(1−sin x)⋅(1−sin x)
x)⋅(1−sin x)
1−sin x
= (
=
) = ( (1−sin1−sin
)
)
1+sin x
(1−sin
x)⋅(1+
sin
x)
x
2
2
(ii) From (I) with x replaced by 2x
(
1
−tan 2 x
cos 2 x
2x
1
1
=
⇒ 1−sin2 x = ⋅(1+sin 2 x) ⇒ 3−3 sin2 x = 1+sin 2 x
) = 13 ⇒ 1−sin
1+sin 2 x
3
3
2
⇒ 2 = 4⋅sin 2 x ⇒ sin 2 x =
1
2
⇒ 2x =
π
5π
or
6
6
⇒
x =
π
5π
or
12
12
205
(i) The range of f (x) is [−1 ,5]
(iii) p= π
x = 2−3 cos y ⇒ 3 cos y = 2−x
⇒
−1
−1
g (x ) = cos
( 2−x3 )
206
(i) 1 = sin2 x+ cos2 x = (a+ b)2+(a−b)2 = a 2+2 ab+ b2 +a 2−2 ab+b 2 = 2⋅(a2 +b 2) ⇒ a 2+ b2 ≡
(ii) 2 = tan x =
sin x
a+b
=
cos x
a−b
⇒ 2 a−2b=a+ b ⇒ a=3 b
1
2
207
(i) Note 2 y +
(3 x)
=5 ⇒
π
5 3x
y line = Y 2= −
, min curve=−1/2, Max curve =5 /2 , min curve=−3 , Max curve =3
2 2π
Approximation to the 4 solutions are shows:
208—part (iii) is subtle! ---------- see third page of solution for easier example.
(i) q≤f (x)≤q+ p
(a) f (x)= p+ q
2:
π 3π
,
4 4
(b) f (x)=q
3 :0 ,
π
,π
2
1
(c) f (x)= p+ q
2
π 3π 5 π 7 π
4 :0 , ,
,
,
8 8 8 8
208—part (iii) is subtle! ---------- see third page of solution for easier example.
4 = f (x) = p⋅sin 2 2 x+ q = 3⋅sin2 2 x+ 2 ⇒ sin2 2 x =
⇒
x =
1
⋅sin−1
2
2
3
2
3
(√ 23 ) ----------or------- x = − 12⋅sin (√ )
−1
2
⇒ sin 2 x =
2
3
⇒ sin 2 x = ±
√
2
3
208—part (iii) is subtle! ---------- see third page of solution for easier example.
An example easier to understand would be
1
=sin(2 x) ⇒ 2 x = 30 ° ,150 ° , 210° , 330 °, … or 30 ° +k⋅180 ° , 150°+ k⋅180 ° ,…
2
⇒
x = 15 ° +k⋅90° , 75 °+ k⋅90 ° , …
209
(i) 0 = 4 tan x +3 cos x +
1
x
cos
⇒ 0 = 4 sin x+3 cos 2 x +1 ⇒ 0 = 4 sin x+3−3 sin2 x+1
⇒ 0 = 3 sin2 x−4 sin x −4 = (sin x−2)⋅(3sin x +2) ⇒ sin x = −
2
[since sin x cannot be bigger than 1]
3
(ii) Using (i) with x replaced by 2 x−20 ° yields
sin(2 x−20 ° ) = −
⇒
2
3
x ≈ 169.1°, 120.9°
⇒ 2 x−20 ° = sin
−1
(− 23 ) ⇒ 2 x−20 ° ≈ 318.2°, 221.8°
210
−1
(a) 1 = 3 tan (2 x +1) ⇒ (a) 2 x+1 = tan
≈ 0.32, 3.46, 6.60, … ⇒
( 13 ) ≈ 0.32+m⋅π = 0.32, 0.32 + π, 0.32 + 2π, ...
x ≈ -0.34, 1.23, 2.80, ...
210 continued
(i) f (x) = 3 cos 2 x−2 sin2 x = 3 cos 2 x−2⋅(1−cos 2 x ) = 5 cos 2 x−2
(ii) [−2≤f (x)≤3]
These graphs show Y 1=3 cos 2 x−2sin 2 x = 5 cos 2 x−2 to be Y 2=5 cos 2 x shifted down by 2.
x =
[ ( ) ]
1
2
⋅ sin−1 − +20 °
2
3
° π θ α β γ
ʃ √ ≅ ≡ ≈ ≠ ⇒
6⋅cos θ ⋅tanθ −3⋅cos θ + 4 tanθ −2 ⇔
cos−1
√
3
()
1
2
a
b
Pshaded = |^
BD|+|DC|+|CB|
−b±√ b2−4 ac
2a